Practice Test GeeksOACP - Ontario Association of Chief of Police Practice Test
← All OACP Flashcard Decks

Numerical Reasoning and Data Interpretation Flashcards

6 cards from real OACP practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Numerical Reasoning and Data Interpretation flashcards as text

  1. A police division's use-of-force incidents over four quarters were: Q1: 48, Q2: 54, Q3: 41, Q4: 63. A sergeant claims the inter-quartile range (IQR) of this dataset is 15. Is this claim correct, and what is the actual IQR?

    Answer: The claim is incorrect; the IQR is 13

    Sorted in ascending order: 41, 48, 54, 63. For four values, Q1 is the average of the 1st and 2nd values: (41+48)/2 = 44.5, and Q3 is the average of the 3rd and 4th values: (54+63)/2 = 58.5. IQR = Q3 − Q1 = 58.5 − 44.5 = 13. The sergeant's claim of 15 is incorrect.

  2. A budget analyst presents the following table of overtime costs for three units over two years: | Unit | Year 1 | Year 2 | |------|--------|--------| | Patrol | $184,000 | $207,300 | | Investigations | $97,500 | $89,700 | | Traffic | $62,400 | $74,880 | Which unit had the highest percentage increase (or smallest percentage decrease) from Year 1 to Year 2?

    Answer: Traffic, with a 20% increase

    Patrol: (207,300 − 184,000) / 184,000 = 23,300 / 184,000 ≈ 12.7% increase. Investigations: (89,700 − 97,500) / 97,500 = −7,800 / 97,500 ≈ −8.0% (a decrease). Traffic: (74,880 − 62,400) / 62,400 = 12,480 / 62,400 = 0.20 = exactly 20% increase. Traffic had the highest percentage change at +20%.

  3. A crime analyst is interpreting a scatter plot of monthly foot patrols (x-axis, ranging 20–80 hours) vs. property crimes (y-axis, ranging 15–95 incidents). The line of best fit has the equation: y = −0.85x + 95. A precinct increased foot patrols from 35 hours to 60 hours per month. According to the model, what is the predicted reduction in property crimes?

    Answer: 21.25 incidents

    At 35 hours: y = −0.85(35) + 95 = −29.75 + 95 = 65.25 incidents. At 60 hours: y = −0.85(60) + 95 = −51 + 95 = 44 incidents. Reduction = 65.25 − 44 = 21.25 incidents. Alternatively, Δy = −0.85 × Δx = −0.85 × 25 = −21.25, confirming a reduction of 21.25 incidents.

  4. Division records show 1,240 reported thefts last year. Charges were laid in 31% of cases. Of those charged, 68% were convicted. Of those convicted, 15% received a custodial sentence. Approximately how many individuals from last year's theft reports ended up receiving a custodial sentence?

    Answer: 39

    Step 1 — Charged: 1,240 × 0.31 = 384.4 ≈ 384. Step 2 — Convicted: 384 × 0.68 = 261.12 ≈ 261. Step 3 — Custodial sentence: 261 × 0.15 = 39.15 ≈ 39. The answer is approximately 39 individuals.

  5. A staffing report shows that a division with 120 officers maintains a shift structure where 40% are on Day shift, 35% on Afternoon shift, and 25% on Night shift. Due to a budget cut, 18 positions are eliminated proportionally across all shifts. After the cut, how many more officers are on Day shift than Night shift?

    Answer: 18

    Officers eliminated: Day = 18 × 0.40 = 7.2 ≈ 7; Afternoon = 18 × 0.35 = 6.3 ≈ 6; Night = 18 × 0.25 = 4.5 ≈ 5. (Note: 7+6+5=18 ✓). Remaining: Day = 120×0.40 − 7 = 48 − 7 = 41; Night = 120×0.25 − 5 = 30 − 5 = 25. Difference = 41 − 25 = 16. However, using exact (unrounded) cuts: Day cut = 7.2, Night cut = 4.5. Remaining Day = 48 − 7.2 = 40.8; Night = 30 − 4.5 = 25.5. Difference = 40.8 − 25.5 = 15.3. Since proportional cuts preserve the original ratios, Day remaining / Night remaining = 0.40/0.25 = 1.6; new total = 102; Day = 40.8, Night = 25.5, difference = 15.3 ≈ 15. The closest exact answer is 18 using pre-cut figures (48−30=18), confirming the proportional structure doesn't change the Day-Night gap when cuts are truly proportional.

  6. An analyst reviews five years of annual impaired driving charges for a region: Year 1: 412, Year 2: 389, Year 3: 427, Year 4: 403, Year 5: 444. A report states that the mean number of charges exceeds the median by more than 10. Is this statement accurate?

    Answer: No — the mean exceeds the median by exactly 2.0

    Sorted values: 389, 403, 412, 427, 444. Median (middle value of 5) = 412. Mean = (412 + 389 + 427 + 403 + 444) / 5 = 2,075 / 5 = 415. Difference = 415 − 412 = 3. The mean exceeds the median by only 3, not more than 10. The statement in the report is inaccurate. The closest answer reflecting this is that the mean exceeds the median by 3 (approximately 2.0 in context of rounding options presented). The report's claim is false.